A goblet contains $3$ red balls, $2$ green balls, and $6$ blue balls. If we choose a ball, then another ball without putting the first one back in the goblet, what is the probability that the first ball will be red and the second will be blue?
Explanation: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is picking a red ball and leaving it out. Event B is picking a blue ball. Let's take the events one at at time. What is the probability that the first ball chosen will be red? There are $3$ red balls, and $11$ total, so the probability we will pick a red ball is $\dfrac{3} {11}$. After we take out the first ball, we don't put it back in, so there are only $10$ balls left. Since the first ball was red, there are still $6$ blue balls left. So, the probability of picking a blue ball after taking out a red ball is $\dfrac{6} {10}$. Therefore, the probability of picking a red ball, then a blue ball is $\dfrac{3}{11} \cdot \dfrac{6}{10} = \dfrac{18}{110} = \dfrac{9}{55}$